Partially Filled Horizontal Tank Volume with 2:1 Elliptical Heads

\[V=L\left(\frac{D^2}{4}\arccos\left(\frac{D-2h}{D}\right)-\left(\frac{D}{2}-h\right)\sqrt{Dh-h^2}\right)+\frac{\pi h^2(3D-2h)}{12}\]

Variables

Vliquid volume in the partially filled tank (m³)

Dinternal tank diameter (m)

Linternal cylindrical shell length (m)

hliquid depth measured from tank bottom (m)

Description

What is this formula?

Partially Filled Horizontal Tank Volume with 2:1 Elliptical Heads calculates the liquid volume inside a horizontal cylindrical tank with two standard 2:1 elliptical heads.

The formula combines the liquid volume in the cylindrical shell with the partial volume contained in the two elliptical heads.

For a 2:1 elliptical head, the head depth is D/4, which allows the formula to use only D, L, and h.

When to use it

Use this formula when the tank is horizontal, has two 2:1 elliptical heads, and is only partially filled.

It is useful for tank gauging, inventory estimation, process vessels, chemical tanks, and pressure vessels.

Example

A horizontal tank with 2:1 elliptical heads has:

D = 2.0 m

L = 10.0 m

h = 0.80 m

Formula:

V = L × [(D²/4) × arccos((D−2h)/D) − ((D/2)−h) × √(Dh−h²)] + πh²(3D−2h)/12

Substitution:

V = 10 × [(2²/4) × arccos((2−2×0.8)/2) − ((2/2)−0.8) × √(2×0.8−0.8²)] + π×0.8²×(3×2−2×0.8)/12

Result:

V = 12.47 m³

The partially filled tank contains approximately 12.47 m³ of liquid.

Applications

- 2:1 elliptical head tank gauging

- Horizontal pressure vessel inventory

- Chemical storage tanks

- Petroleum process vessels

- Industrial liquid level calculations

- Tank capacity estimation

Note

This formula assumes two identical 2:1 elliptical heads, where each head depth equals D/4.

It is valid for liquid depths from 0 to D.

For general ellipsoidal heads with a different head depth, use the partially filled horizontal tank volume with ellipsoidal heads formula instead.